# 8.5: Pressure Dependence of Gibbs Energy

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- Contributed by Patrick Fleming
- Assistant Professor (Chemistry and Biochemistry) at California State University East Bay

The pressure and temperature dependence of \(G\) is also easy to describe. The best starting place is the definition of \(G\).

\[G = U + pV -TS \label{eq1} \]

Taking the total differential of \(G\) yields

\[dG = dU + pdV – pdV + Vdp – TdS – SdT\]

The differential can be simplified by substituting the combined first and second law statement for \(dU\) (consider a reversible process and \(pV\) work only).

\[dG = \cancel{TdS} \bcancel{– pdV} + \bcancel{pdV} + Vdp – \cancel{TdS} – SdT\]

Canceling the \(TdS\) and \(pdV\) terms leaves

\[dG = V\,dp – S\,dT \label{Total1}\]

This suggests that the natural variables of \(G\) are \(p\) and \(T\). So the total differential \(dG\) can also be expressed

\[ dG = \left( \dfrac{\partial G}{\partial p} \right)_T dp + \left( \dfrac{\partial G}{\partial T} \right)_p dT \label{Total2}\]

And by inspection of Equations \ref{Total1} and \ref{Total2}, it is clear that

\[\left( \dfrac{\partial G}{\partial p} \right)_T = V\]

and

\[ \left( \dfrac{\partial G}{\partial T} \right)_p = -S\]

It is also clear that the Maxwell relation on \(G\) is given by

\[\left( \dfrac{\partial V}{\partial T} \right)_p = \left( \dfrac{\partial S}{\partial p} \right)_T\]

which is an extraordinarily useful relationship, since one of the terms is expressible entirely in terms of measurable quantities!

\[\left( \dfrac{\partial V}{\partial T} \right)_p = V\alpha\]

The pressure dependence of \(G\) is given by the pressure derivative at constant temperature

\[\left( \dfrac{\partial G}{\partial p} \right)_T = V \label{Max2}\]

which is simply the molar volume. For a fairly incompressible substance (such as a liquid or a solid) the molar volume will be essentially constant over a modest pressure range.

Example \(\PageIndex{1}\): Gold under Pressure

The density of gold is 19.32 g/cm^{3}. Calculate \(\Delta G\) for a 1.00 g sample of gold when the pressure on it is increased from 1.00 atm to 2.00 atm.

**Solution**:

The change in the Gibbs function due to an isothermal change in pressure can be expressed as

\[ \Delta G =\int_{p_1}^{p_2} \left( \dfrac{\partial G}{\partial p} \right)_T dp\]

And since substituting Equation \ref{Max2}, results in

\[ \Delta G =\int_{p_1}^{p_2} V dp\]

Assuming that the molar volume is independent or pressure over the stated pressure range, \(\Delta G\) becomes

\[\Delta G = V(p_2-p_1)\]

So, the molar change in the Gibbs function can be calculated by substituting the relevant values.

\[ \begin{align} \Delta G & = \left( \dfrac{197.0\, g}{mol} \times \dfrac{1\,}{19.32\,g} \times \dfrac{1\,L}{1000\,cm^3} \right) (2.00 \,atm -1.00 \,atm) \underbrace{ \left(\dfrac{8.315 \,J}{0.08206\, atm\,L}\right)}_{\text{conversion unit}}\\ &= 1.033\,J \end{align}\]

Contributors

Patrick E. Fleming (Department of Chemistry and Biochemistry; California State University, East Bay)